An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V2) Cambridge International A Level Chemistry paper. Not affiliated with or reproduced from Cambridge.
Paper 1CR
Answer all questions. Show all the steps in any calculations and state the units. Calculators and rulers are required.
11 題目 · 110 分
題目 1 · Structured/Short Answer
10 分
A student heats a sample of hydrated iron(II) sulfate, \(\text{FeSO}_4 \cdot x\text{H}_2\text{O}\), in a crucible to determine the value of \(x\).
(a) Write a chemical equation for the thermal decomposition of hydrated iron(II) sulfate to form anhydrous iron(II) sulfate and water. [1]
(b) In the experiment, the mass of the empty crucible was \(18.60\text{ g}\). The mass of the crucible and hydrated iron(II) sulfate was \(24.16\text{ g}\). After heating to constant mass, the mass of the crucible and anhydrous iron(II) sulfate was \(21.64\text{ g}\).
(i) Calculate the mass of anhydrous iron(II) sulfate formed and the mass of water lost. [2]
(ii) Calculate the number of moles of anhydrous iron(II) sulfate (\(\text{FeSO}_4\)) and water (\(\text{H}_2\text{O}\)) in the sample. (\(M_r\text{ of FeSO}_4 = 152\), \(M_r\text{ of H}_2\text{O} = 18\)). [3]
(iii) Determine the value of \(x\) in \(\text{FeSO}_4 \cdot x\text{H}_2\text{O}\). Show your working. [2]
(c) Suggest why the student heats the crucible to 'constant mass'. [2]
(b)(i) Mass of anhydrous iron(II) sulfate = \(21.64\text{ g} - 18.60\text{ g} = 3.04\text{ g}\). Mass of water lost = \(24.16\text{ g} - 21.64\text{ g} = 2.52\text{ g}\).
(b)(iii) Ratio of moles \(\text{H}_2\text{O} : \text{FeSO}_4 = \frac{0.140\text{ mol}}{0.020\text{ mol}} = 7\). Therefore, the value of \(x\) is 7.
(c) To ensure that all of the water of crystallization has been completely driven off / lost from the hydrated salt.
評分準則
(a) 1 Mark: Correct reactants and products (\(\text{FeSO}_4\cdot x\text{H}_2\text{O} \rightarrow \text{FeSO}_4 + x\text{H}_2\text{O}\)). State symbols are not required.
(b)(i) 2 Marks: 1 mark for mass of anhydrous salt (3.04 g) and 1 mark for mass of water lost (2.52 g).
(b)(ii) 3 Marks: 1 mark for calculating moles of \(\text{FeSO}_4\) (0.020 mol), 1 mark for calculating moles of \(\text{H}_2\text{O}\) (0.140 mol), 1 mark for correct working shown for both calculations.
(b)(iii) 2 Marks: 1 mark for finding the simplest whole-number ratio (\(0.140 / 0.020 = 7\)), 1 mark for correct integer value \(x = 7\).
(c) 2 Marks: 1 mark for mentioning 'to ensure all water has evaporated/been lost', 1 mark for explaining that heating is repeated until there is no further change in mass.
題目 2 · Structured/Short Answer
10 分
A student investigates the temperature change during the displacement reaction between zinc powder and copper(II) sulfate solution.
(a) The student adds \(50.0\text{ cm}^3\) of \(0.400\text{ mol/dm}^3\) copper(II) sulfate solution to a polystyrene cup. The initial temperature is \(19.5 ^\circ\text{C}\). After adding an excess of zinc powder and stirring, the maximum temperature reached is \(34.2 ^\circ\text{C}\).
(i) Show that the heat energy released, \(q\), is approximately \(3080\text{ J}\). (Assume the density of the solution is \(1.00\text{ g/cm}^3\) and the specific heat capacity is \(4.18\text{ J/g/}^\circ\text{C}\)). [3]
(ii) Calculate the number of moles of copper(II) sulfate used in the reaction. [1]
(iii) Calculate the molar enthalpy change (\(\Delta H\)) for the reaction in \(\text{kJ/mol}\). Include a sign in your answer. [3]
(b) Draw an energy level diagram for this reaction, showing the reactants, products, and \(\Delta H\). [3]
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解題
(a)(i) Temperature change \(\Delta T = 34.2 - 19.5 = 14.7 ^\circ\text{C}\). Mass of solution \(m = 50.0\text{ cm}^3 \times 1.00\text{ g/cm}^3 = 50.0\text{ g}\). Heat energy released, \(q = m c \Delta T = 50.0\text{ g} \times 4.18\text{ J/g/}^\circ\text{C} \times 14.7 ^\circ\text{C} = 3072.3\text{ J}\), which is approximately \(3080\text{ J}\).
(a)(ii) \(\text{Moles of CuSO}_4 = \text{volume in dm}^3 \times \text{concentration} = \frac{50.0}{1000} \times 0.400 = 0.0200\text{ mol}\).
(a)(iii) \(\Delta H = -\frac{q}{\text{moles}} = -\frac{3072.3\text{ J}}{0.0200\text{ mol}} = -153615\text{ J/mol} = -153.6\text{ kJ/mol}\). (Using the rounded 3080 J: \(\Delta H = -\frac{3080\text{ J}}{0.0200\text{ mol}} = -154\text{ kJ/mol}\)).
(b) An energy level diagram must be drawn: a horizontal line representing the reactants (\(\text{Zn} + \text{CuSO}_4\)) is drawn at a higher level than a horizontal line representing the products (\(\text{ZnSO}_4 + \text{Cu}\)). A vertical downward arrow from reactants to products represents \(\Delta H\), with a negative value.
評分準則
(a)(i) 3 Marks: 1 mark for finding \(\Delta T = 14.7 ^\circ\text{C}\); 1 mark for substituting values into \(q = mc\Delta T\); 1 mark for calculating correct value of 3072.3 J.
(a)(iii) 3 Marks: 1 mark for converting J to kJ; 1 mark for dividing energy by moles; 1 mark for correct final answer with a negative sign (-153.6 or -154 kJ/mol).
(b) 3 Marks: 1 mark for energy/enthalpy axis and showing reactants at a higher energy than products; 1 mark for labeling reactants (\(\text{Zn} + \text{CuSO}_4\)) and products (\(\text{ZnSO}_4 + \text{Cu}\)); 1 mark for downward arrow labeled with negative value of \(\Delta H\).
題目 3 · Structured/Short Answer
10 分
Hydrogen peroxide decomposes slowly at room temperature to form water and oxygen gas according to the equation: \(2\text{H}_2\text{O}_2(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})\)
(a) (i) Describe the apparatus a student could use to measure the volume of gas produced over time during this reaction. [3]
(ii) State the effect of adding manganese(IV) oxide catalyst on the rate of this reaction. [1]
(iii) Explain, in terms of collision theory and activation energy, how a catalyst increases the rate of reaction. [3]
(b) The student repeats the experiment at a higher temperature, keeping all other variables constant. Explain, in terms of particles, why the rate of reaction increases when the temperature is increased. [3]
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解題
(a)(i) A reaction vessel (conical flask) containing hydrogen peroxide is sealed with a stopper. A delivery tube connects the flask to a gas syringe or to an inverted measuring cylinder filled with water. A stopwatch is used to record the volume of oxygen gas produced at regular time intervals.
(a)(ii) The rate of reaction increases (the reaction becomes faster).
(a)(iii) A catalyst provides an alternative reaction pathway with a lower activation energy. Consequently, a larger proportion of colliding particles have energy equal to or greater than the activation energy, leading to an increased frequency of successful collisions.
(b) At a higher temperature, particles have more kinetic energy and move faster, leading to a higher frequency of collisions. More importantly, a much higher proportion of particles possess energy greater than or equal to the activation energy, so a larger fraction of collisions are successful, significantly increasing the rate of reaction.
評分準則
(a)(i) 3 Marks: 1 mark for conical flask containing reactants; 1 mark for delivery tube; 1 mark for gas syringe / inverted measuring cylinder over water to collect gas.
(a)(ii) 1 Mark: increases rate of reaction.
(a)(iii) 3 Marks: 1 mark for 'alternative reaction pathway'; 1 mark for 'lower activation energy'; 1 mark for 'increased frequency of successful collisions'.
(b) 3 Marks: 1 mark for 'particles have more kinetic energy / move faster'; 1 mark for 'more frequent collisions'; 1 mark for 'greater proportion of particles have energy \(\ge\) activation energy / more successful collisions per unit time'.
題目 4 · Structured/Short Answer
10 分
A student electrolyzes concentrated sodium chloride solution (brine) using inert carbon electrodes.
(a) State the name of the gas produced at: (i) the anode (positive electrode) [1] (ii) the cathode (negative electrode) [1]
(b) Write ionic half-equations, including state symbols, for the reactions occurring at: (i) the anode [2] (ii) the cathode [2]
(c) After the electrolysis has run for some time, a few drops of universal indicator are added to the solution around the cathode. (i) State the color change observed and the pH range of the solution at this electrode. [2] (ii) Explain why this change in pH occurs. [2]
(c)(i) The solution turns blue or purple, indicating a pH of 11-14 (alkaline). (c)(ii) Hydrogen ions are discharged at the cathode, leaving behind hydroxide ions (\(\text{OH}^-\)) from the self-ionization of water. The build-up of sodium (\(\text{Na}^+\)) and hydroxide (\(\text{OH}^-\)) ions creates a highly alkaline sodium hydroxide solution.
(b)(i) 2 Marks: 1 mark for correct reactants and products (\(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\)); 1 mark for correct state symbols.
(b)(ii) 2 Marks: 1 mark for correct reactants and products (\(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\)); 1 mark for correct state symbols.
(c)(i) 2 Marks: 1 mark for color change (turns blue or purple); 1 mark for pH range (11 to 14).
(c)(ii) 2 Marks: 1 mark for explaining that hydrogen ions are discharged/removed; 1 mark for explaining that hydroxide (\(\text{OH}^-\)) ions remain in excess.
題目 5 · Structured/Short Answer
10 分
A student is given a green solid, Compound X, which is a soluble transition metal salt.
(a) The student dissolves X in deionized water to form a green solution, then adds sodium hydroxide solution. A green precipitate forms which is insoluble in excess sodium hydroxide. (i) Identify the cation present in X. [1] (ii) Write the ionic equation, including state symbols, for the formation of this green precipitate. [2]
(b) To another portion of the solution of X, the student adds dilute hydrochloric acid followed by barium chloride solution. A white precipitate forms. (i) Identify the anion present in X. [1] (ii) Write the ionic equation, including state symbols, for the formation of this white precipitate. [2]
(c) State the chemical formula of Compound X. [1]
(d) Describe a chemical test, and its positive result, that can be used to show that a gas is carbon dioxide. [3]
(d) Bubble the gas into limewater (calcium hydroxide solution). The limewater will turn cloudy or milky if carbon dioxide is present.
評分準則
(a)(i) 1 Mark: iron(II) / \(\text{Fe}^{2+}\). (a)(ii) 2 Marks: 1 mark for correct species and balancing; 1 mark for correct state symbols.
(b)(i) 1 Mark: sulfate / \(\text{SO}_4^{2-}\). (b)(ii) 2 Marks: 1 mark for correct species and balancing; 1 mark for correct state symbols.
(c) 1 Mark: \(\text{FeSO}_4\) (reject iron sulfate / incorrect formulas).
(d) 3 Marks: 1 mark for bubble gas into limewater; 1 mark for cloudy/milky appearance; 1 mark for specifying the test reagent is 'limewater'.
題目 6 · Structured/Short Answer
10 分
A student investigates the reactivity of four metals: magnesium, zinc, iron, and copper by reacting zinc with solutions of metal nitrates.
(a) The student records the following observations: - Zinc + Magnesium nitrate: No reaction - Zinc + Copper(II) nitrate: Red-brown solid forms, blue solution turns colorless - Zinc + Iron(II) nitrate: Grey solid forms, pale green solution turns colorless
(i) Place the four metals in order of their reactivity, from most reactive to least reactive. [2]
(ii) Write a chemical equation for the displacement reaction that occurs between zinc and copper(II) nitrate. [2]
(iii) Explain, in terms of electrons, why the reaction between zinc and copper(II) ions is a redox reaction. [3]
(b) Explain why magnesium cannot be extracted from its oxide ore by heating with carbon, whereas iron can be. [3]
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解題
(a)(i) Magnesium, Zinc, Iron, Copper (from most to least reactive).
(a)(iii) Zinc (\(\text{Zn}\)) is oxidized because it loses electrons (\(\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-\)). Copper(II) ions (\(\text{Cu}^{2+}\)) are reduced because they gain electrons (\(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\)). Since both oxidation and reduction occur in the same reaction, it is a redox reaction.
(b) Magnesium is more reactive than carbon, meaning carbon is not strong enough to remove oxygen from magnesium oxide (carbon cannot reduce it). Iron is less reactive than carbon, so carbon can readily displace/reduce iron from iron oxides.
評分準則
(a)(i) 2 Marks: 2 marks for all four in the correct order. 1 mark if one swap occurs.
(a)(ii) 2 Marks: 1 mark for correct reactants and products; 1 mark for balancing.
(a)(iii) 3 Marks: 1 mark for explaining zinc loses electrons / undergoes oxidation; 1 mark for explaining copper(II) ions gain electrons / undergo reduction; 1 mark for linking oxidation and reduction occurring simultaneously to redox.
(b) 3 Marks: 1 mark for comparing reactivity of magnesium and carbon; 1 mark for comparing reactivity of iron and carbon; 1 mark for stating that carbon can only reduce oxides of less reactive metals.
題目 7 · Structured/Short Answer
10 分
This question is about Group 1 elements.
(a) A teacher adds a small piece of sodium to a trough of water containing a few drops of phenolphthalein indicator. (i) Describe three physical observations that can be made during this reaction. [3] (ii) Write a balanced chemical equation, including state symbols, for the reaction of sodium with water. [3]
(b) Rubidium (Rb) is located below potassium (K) in Group 1. (i) Predict how the reactivity of rubidium with water compares to that of potassium. [1] (ii) Explain this difference in reactivity in terms of the electronic configurations of their atoms. [3]
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解題
(a)(i) Observations include: the sodium floats on the surface of the water, it melts into a spherical ball, it moves rapidly across the surface, there is effervescence (fizzing), and the indicator turns pink.
(b)(ii) A rubidium atom has more occupied electron shells than a potassium atom. This means its single outer shell electron is further from the positive nucleus, and is shielded by more inner shells. Therefore, the electrostatic attraction between the nucleus and the outer electron is weaker, making it easier for rubidium to lose its outer electron and react.
評分準則
(a)(i) 3 Marks: 1 mark for each of any three distinct valid observations. Max 3 marks.
(a)(ii) 3 Marks: 1 mark for correct reactants and products; 1 mark for correct balancing; 1 mark for correct state symbols.
(b)(i) 1 Mark: Correctly predicting rubidium is more reactive.
(b)(ii) 3 Marks: 1 mark for stating rubidium has more electron shells (further from nucleus); 1 mark for weaker electrostatic attraction between the nucleus and outer electron; 1 mark for stating the outer electron is lost more easily.
題目 8 · Structured/Short Answer
10 分
Propene (\(\text{C}_3\text{H}_6\)) is an unsaturated hydrocarbon that belongs to the homologous series of alkenes.
(a) (i) Draw the displayed formula of propene. [1] (ii) Define the term 'unsaturated hydrocarbon'. [2]
(b) Propene reacts with bromine water. (i) State the color change observed when bromine water is shaken with propene. [2] (ii) Draw the displayed formula of the product formed when propene reacts with bromine. [2]
(c) Propene can undergo addition polymerization to form poly(propene). (i) Write an equation for this polymerization reaction using displayed formulas. [2] (ii) State one environmental problem associated with the disposal of addition polymers like poly(propene). [1]
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解題
(a)(i) Propene has the structural formula \(\text{CH}_2=\text{CH-CH}_3\). The displayed formula must show every single C-H, C-C, and C=C bond explicitly.
(a)(ii) A 'hydrocarbon' is a compound consisting of hydrogen and carbon atoms only. 'Unsaturated' means that the molecule contains carbon-to-carbon double bonds (\(\text{C}=\text{C}\)).
(b)(i) The bromine water changes color from orange/brown to colorless.
(b)(ii) The product of this addition reaction is 1,2-dibromopropane, which contains a single C-C bond in place of the C=C bond, and a bromine atom attached to each of the first two carbons.
(c)(i) The polymerization equation shows \(n\) repeating monomer units (propene) with a C=C double bond yielding a repeating polymer chain unit in square brackets with single bonds extending beyond the brackets and a subscript \(n\).
(c)(ii) Addition polymers are inert and non-biodegradable, meaning they do not break down naturally in landfill sites, or they release toxic/greenhouse gases when burned.
評分準則
(a)(i) 1 Mark: Correct displayed structure of propene showing all bonds, including the double bond and all C-H bonds.
(a)(ii) 2 Marks: 1 mark for 'hydrogen and carbon only'; 1 mark for 'contains carbon-carbon double bonds'.
(b)(i) 2 Marks: 1 mark for starting color (orange/brown/yellow); 1 mark for final color (colorless - reject clear).
(b)(ii) 2 Marks: 1 mark for correct single carbon-carbon bonds and overall layout; 1 mark for two bromine atoms added to adjacent carbons.
(c)(i) 2 Marks: 1 mark for reactant with \(n\) in front of propene monomer; 1 mark for correct repeating unit in square brackets with subscript \(n\) and open-ended single bonds.
(c)(ii) 1 Mark: Mentioning that it is non-biodegradable (or takes up landfill space, or produces toxic gases when burned).
題目 9 · Structured/Short Answer
10 分
A student carries out an experiment to determine the formula of hydrated magnesium sulfate, \(\text{MgSO}_4 \cdot x\text{H}_2\text{O}\).
(a) Describe how the student could heat a sample of hydrated magnesium sulfate in a crucible to determine the mass of water lost. Include how they ensure all the water has been removed. (4 marks)
(b) The student obtains the following results from the experiment: - Mass of empty crucible = \(25.40\text{ g}\) - Mass of crucible + hydrated magnesium sulfate = \(30.32\text{ g}\) - Mass of crucible + anhydrous magnesium sulfate = \(27.80\text{ g}\)
(i) Calculate the mass of anhydrous magnesium sulfate formed. (1 mark)
(ii) Calculate the mass of water lost. (1 mark)
(iii) Calculate the value of \(x\) in \(\text{MgSO}_4 \cdot x\text{H}_2\text{O}\). Show all your working. (Relative formula masses, \(M_{\text{r}}\): \(\text{MgSO}_4 = 120\), \(\text{H}_2\text{O} = 18\)). (4 marks)
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解題
(a) To determine the mass of water lost: 1. Weigh the empty crucible, then weigh the crucible containing the hydrated magnesium sulfate. 2. Heat the crucible and its contents strongly using a Bunsen burner (resting on a pipeclay triangle and tripod) for several minutes. 3. Allow the crucible to cool and reweigh. 4. Heat, cool, and reweigh again until a constant mass is reached (this ensures all the water of crystallisation has been driven off).
(b) Calculations: (i) Mass of anhydrous \(\text{MgSO}_4 = 27.80\text{ g} - 25.40\text{ g} = 2.40\text{ g}\)
(ii) Mass of water lost = \(30.32\text{ g} - 27.80\text{ g} = 2.52\text{ g}\) (or \(4.92\text{ g} - 2.40\text{ g} = 2.52\text{ g}\))
(iii) Calculation of \(x\): - Moles of anhydrous \(\text{MgSO}_4 = \frac{2.40\text{ g}}{120\text{ g/mol}} = 0.0200\text{ mol}\) - Moles of water, \(\text{H}_2\text{O} = \frac{2.52\text{ g}}{18\text{ g/mol}} = 0.140\text{ mol}\) - Mole ratio of \(\text{H}_2\text{O} : \text{MgSO}_4 = \frac{0.140}{0.0200} = 7\) - Therefore, \(x = 7\).
評分準則
**Part (a) [4 marks]** - M1: Weigh the empty crucible and weigh crucible with sample / record initial masses [1] - M2: Heat the crucible using a Bunsen burner [1] - M3: Cool and reweigh [1] - M4: Repeat heating and weighing until a constant mass is obtained (to ensure all water is lost) [1]
**Part (b)(i) [1 mark]** - \(2.40\text{ g}\) (correct calculation of anhydrous salt mass) [1]
**Part (b)(ii) [1 mark]** - \(2.52\text{ g}\) (correct calculation of water mass lost) [1]
**Part (b)(iii) [4 marks]** - M1: Calculate moles of \(\text{MgSO}_4 = 0.0200\text{ mol}\) [1] (allow ecf from b(i)) - M2: Calculate moles of \(\text{H}_2\text{O} = 0.140\text{ mol}\) [1] (allow ecf from b(ii)) - M3: Divide both by the smaller number to get ratio (\(0.140 / 0.0200 = 7\)) [1] - M4: State \(x = 7\) [1] (must be a whole number; ecf allowed only if working is shown)
題目 10 · Structured/Short Answer
10 分
A student uses a calorimetry experiment to investigate the temperature change of the displacement reaction between zinc and copper(II) sulfate.
They place \(50.0\text{ cm}^3\) of \(0.400\text{ mol/dm}^3\) copper(II) sulfate solution into a polystyrene cup. They record the initial temperature of the solution, then add an excess of zinc powder and stir. They record the maximum temperature reached. - Initial temperature = \(19.5\text{ }^\circ\text{C}\) - Maximum temperature = \(31.8\text{ }^\circ\text{C}\)
(a) Explain why a polystyrene cup is used instead of a glass beaker. (2 marks)
(b) Show by calculation that the heat energy change (\(q\)) in this reaction is approximately \(2570\text{ J}\). (Assume the density of the solution is \(1.00\text{ g/cm}^3\), and its specific heat capacity is \(4.18\text{ J/g/}^\circ\text{C}\).) (3 marks)
(c) Calculate the molar enthalpy change (\(\Delta H\)) for the reaction in \(\text{kJ/mol}\). Include a sign in your final answer. (4 marks)
(d) State whether this reaction is exothermic or endothermic, giving a reason from the results. (1 mark)
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解題
(a) Polystyrene is a good thermal insulator [1], which reduces heat loss to the surroundings, leading to a more accurate temperature change measurement [1].
(b) Calculate temperature change: \(\Delta T = 31.8\text{ }^\circ\text{C} - 19.5\text{ }^\circ\text{C} = 12.3\text{ }^\circ\text{C}\)
Calculate mass of solution: \(m = 50.0\text{ cm}^3 \times 1.00\text{ g/cm}^3 = 50.0\text{ g}\)
Using the formula \(q = mc\Delta T\): \(q = 50.0\text{ g} \times 4.18\text{ J/g/}^\circ\text{C} \times 12.3\text{ }^\circ\text{C} = 2570.7\text{ J}\) This is approximately \(2570\text{ J}\).
Calculate molar enthalpy change, \(\Delta H\): \(\Delta H = -\frac{q}{\text{moles}}\) \(\Delta H = -\frac{2570.7\text{ J}}{0.0200\text{ mol}} = -128,535\text{ J/mol}\) Convert to \(\text{kJ/mol}\): \(\Delta H = -128.5\text{ kJ/mol}\) (or \(-129\text{ kJ/mol}\) using \(2570\text{ J}\)).
(d) The reaction is exothermic because the temperature of the reaction mixture increased / heat is given out to the surroundings [1].
評分準則
**Part (a) [2 marks]** - M1: Polystyrene is a good thermal insulator / poor conductor of heat [1] - M2: This reduces heat loss to the surroundings / makes the temperature change measured more accurate [1]
**Part (b) [3 marks]** - M1: Calculate temperature rise, \(\Delta T = 12.3\text{ }^\circ\text{C}\) [1] - M2: Recall and use \(q = mc\Delta T\) with mass = \(50.0\text{ g}\) [1] - M3: Obtain final value of \(2570.7\text{ J}\) (or \(2570\text{ J}\)) [1]
**Part (c) [4 marks]** - M1: Calculate moles of \(\text{CuSO}_4 = 0.0200\text{ mol}\) [1] - M2: State that zinc is in excess, so copper(II) sulfate is limiting [1] - M3: Divide \(q\) in kJ by moles: \(\frac{2.57}{0.0200}\) [1] - M4: Value is \(-128.5\text{ kJ/mol}\) (accept \(-128\) to \(-129\text{ kJ/mol}\)) with a negative sign [1]
**Part (d) [1 mark]** - Exothermic because the temperature increased [1]
題目 11 · Structured/Short Answer
10 分
A student investigates the rate of reaction between marble chips (calcium carbonate, \(\text{CaCO}_3\)) and dilute hydrochloric acid (\(\text{HCl}\)) at room temperature.
They measure the volume of carbon dioxide gas produced over time using a gas syringe.
(a) State two variables that must be kept constant to ensure a fair comparison when comparing the rate of reaction of different sizes of marble chips. (2 marks)
(b) Describe how the rate of reaction changes over time, referencing the slope of a typical volume-time graph. Explain this change in rate in terms of the collision theory. (4 marks)
(c) The student repeats the experiment using the same mass of marble chips, but in a powdered form instead of large chips.
(i) State the effect of this change on the initial rate of reaction. (1 mark)
(ii) Explain this effect in terms of the particle collision theory. (3 marks)
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解題
(a) Control variables include: - Concentration of hydrochloric acid - Volume of hydrochloric acid - Temperature of the reaction mixture - Total mass of calcium carbonate / marble chips (Accept any two of the above)
(b) - Description of rate: The rate of reaction is fastest at the start (as shown by the steepest slope of the graph). As the reaction proceeds, the rate decreases (the slope becomes less steep), and eventually the rate becomes zero / the reaction stops (the graph levels off / becomes flat). - Explanation: The rate decreases because reactant particles (acid particles) are being used up, so their concentration decreases. This leads to a lower frequency of successful collisions between reactant particles per unit time.
(c) (i) The initial rate of reaction increases / the reaction goes faster. (ii) Powdered calcium carbonate has a much larger surface area (to volume ratio) than large chips. This means more calcium carbonate particles are exposed at the surface to react with the acid. Consequently, there is a higher frequency of successful collisions (more successful collisions per unit time).
評分準則
**Part (a) [2 marks]** - Any two control variables from: concentration of acid [1], volume of acid [1], temperature [1], total mass of calcium carbonate [1]. (Reject: amount of acid, size of chips)
**Part (b) [4 marks]** - M1: Rate is fastest at start / slope is steepest, and rate decreases over time / slope becomes less steep [1] - M2: Reaction stops when a reactant is fully used up / curve levels off [1] - M3: Concentration of reactant particles / acid decreases as they are used up [1] - M4: Frequency of successful collisions decreases / fewer successful collisions per unit time [1]
**Part (c)(ii) [3 marks]** - M1: Powder has a larger surface area (to volume ratio) [1] - M2: More reactant particles are exposed to collisions at the surface [1] - M3: Increased frequency of successful collisions / more successful collisions per second / per unit time [1] (Reject: just 'more collisions' without reference to frequency, time, or rate)
Paper 2CR
Answer all questions. Show all the steps in any calculations and state the units. Calculators and rulers are required.
7 題目 · 70 分
題目 1 · structured
10 分
A student investigates the displacement reaction between zinc and copper(II) sulfate solution using a polystyrene cup calorimeter.
(a) Describe how the student should perform this experiment to measure the temperature rise as accurately as possible. (4)
(b) The student adds \(1.50 \text{ g}\) of zinc powder (an excess) to \(50.0 \text{ cm}^3\) of \(0.200 \text{ mol/dm}^3\) copper(II) sulfate solution. The temperature of the mixture increases by \(12.5\ ^\circ\text{C}\).
(i) Calculate the heat energy change (\(q\)) in joules. [Assume the density of the solution is \(1.00 \text{ g/cm}^3\) and its specific heat capacity is \(4.18 \text{ J/g/}^\circ\text{C}\)] (2)
(ii) Calculate the molar enthalpy change (\(\Delta H\)) for the reaction in \(\text{kJ/mol}\). Include a sign in your answer. (3)
(c) Explain why the experimental value of \(\Delta H\) is less exothermic than the theoretical database value. (1)
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解題
(a) Measure \(50.0 \text{ cm}^3\) of copper(II) sulfate solution using a measuring cylinder / volumetric pipette and pour it into a polystyrene cup. Place a thermometer in the solution and record the initial temperature at regular intervals (e.g., every minute) for 3 minutes. At the 4th minute, add \(1.50 \text{ g}\) of zinc powder and stir rapidly. Record the temperature every minute from the 5th to the 10th minute. Plot a graph of temperature against time to find the maximum temperature rise by extrapolation. (b)(i) Mass of solution = \(50.0 \text{ g}\). \(q = m \times c \times \Delta T = 50.0 \times 4.18 \times 12.5 = 2612.5 \text{ J}\). (b)(ii) Amount of \(\text{CuSO}_4 = \text{volume} \times \text{concentration} = 0.0500 \times 0.200 = 0.0100 \text{ mol}\). Since zinc is in excess, \(\text{CuSO}_4\) is the limiting reactant. Enthalpy change = \(-q / \text{moles} = -(2612.5 / 1000) / 0.0100 = -261.25 \text{ kJ/mol}\) (or \(-261 \text{ kJ/mol}\) to 3 s.f.). (c) Heat energy is lost to the surroundings (or absorbed by the polystyrene cup/thermometer).
評分準則
(a) M1: Measure volume of copper(II) sulfate using measuring cylinder / pipette and add to polystyrene cup (1) M2: Record initial temperature at regular intervals before adding zinc (1) M3: Add zinc and stir continuously (1) M4: Record maximum temperature reached or use extrapolation of temperature-time graph (1) (b)(i) M1: Substitution into formula: \(50.0 \times 4.18 \times 12.5\) (1) M2: Correct calculation: \(2612.5\) (J) (1) (b)(ii) M1: Moles of \(\text{CuSO}_4 = 0.0100 \text{ mol}\) (1) M2: Division of heat energy (in kJ) by moles (1) M3: Correct value with negative sign: \(-261\) (kJ/mol) (1) (c) M1: Heat loss to surroundings / heat absorbed by the cup/thermometer (1)
題目 2 · structured
10 分
Sulfur trioxide is produced in the Contact process according to the following reversible reaction:
(a) Explain what is meant by a 'dynamic equilibrium'. (2)
(b) Describe and explain the effect on the position of equilibrium of:
(i) Increasing the pressure. (3)
(ii) Increasing the temperature. (3)
(c) State the catalyst used in this process and explain its effect on the yield of sulfur trioxide. (2)
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解題
(a) Dynamic equilibrium means that the rate of the forward reaction is equal to the rate of the reverse reaction, and the concentrations of reactants and products remain constant in a closed system. (b)(i) Increasing the pressure shifts the position of equilibrium to the right. This is because there are fewer moles of gas on the right-hand side (2 moles) than on the left-hand side (3 moles). The system opposes the increase in pressure by moving in the direction of fewer gas moles. (b)(ii) Increasing the temperature shifts the position of equilibrium to the left. Since the forward reaction is exothermic (releases heat), the system opposes the temperature rise by moving in the endothermic direction (to the left) to absorb heat. (c) The catalyst used is vanadium(V) oxide (\(\text{V}_2\text{O}_5\)). It has no effect on the position of equilibrium or the yield of sulfur trioxide, because it increases the rate of both the forward and reverse reactions equally.
評分準則
(a) M1: Rate of forward reaction equals rate of backward reaction (1) M2: Concentrations of reactants and products remain constant (1) (b)(i) M1: Equilibrium shifts to the right (1) M2: Fewer moles of gas on the right-hand side / 2 moles on right and 3 on left (1) M3: System opposes change by reducing pressure (1) (b)(ii) M1: Equilibrium shifts to the left (1) M2: Forward reaction is exothermic / reverse reaction is endothermic (1) M3: System opposes change by absorbing added heat (1) (c) M1: Vanadium(V) oxide / \(\text{V}_2\text{O}_5\) (1) M2: No effect on yield / position of equilibrium because it increases the rate of forward and backward reactions equally (1)
題目 3 · structured
10 分
The electrolysis of concentrated aqueous sodium chloride (brine) is carried out using inert carbon electrodes.
(a) State the names of the three useful chemical products of this electrolysis process. (3)
(b) Write ionic half-equations, including state symbols, for the reactions occurring at:
(i) the anode (positive electrode). (2)
(ii) the cathode (negative electrode). (2)
(c) Explain why the remaining electrolyte solution gradually becomes alkaline during the process. (2)
(d) Suggest why platinum could be used instead of carbon for the electrodes, but carbon is preferred in industrial cells. (1)
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解題
(a) The three useful products are chlorine gas, hydrogen gas, and sodium hydroxide solution. (b)(i) At the anode (oxidation): \(2\text{Cl}^-\text{(aq)} \rightarrow \text{Cl}_2\text{(g)} + 2\text{e}^-\) (b)(ii) At the cathode (reduction): \(2\text{H}^+\text{(aq)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)}\) (or \(2\text{H}_2\text{O(l)} + 2\text{e}^- \rightarrow \text{H}_2\text{(g)} + 2\text{OH}^-\text{(aq)}\)) (c) Hydrogen ions (\(\text{H}^+\)) and chloride ions (\(\text{Cl}^-\)) are selectively discharged at the electrodes. This leaves sodium ions (\(\text{Na}^+\)) and hydroxide ions (\(\text{OH}^-\)) in the solution, forming sodium hydroxide, which is a strong alkali. (d) Platinum is inert but is much more expensive than carbon.
評分準則
(a) M1: Chlorine (1) M2: Hydrogen (1) M3: Sodium hydroxide (1) (b)(i) M1: \(2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-\) (1) M2: Correct state symbols: \(\text{Cl}^-\text{(aq)}\) and \(\text{Cl}_2\text{(g)}\) (1) (b)(ii) M1: \(2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\) (or equivalent) (1) M2: Correct state symbols: \(\text{H}^-\text{(aq)}\) and \(\text{H}_2\text{(g)}\) (1) (c) M1: \(\text{H}^+\) and \(\text{Cl}^-\) are discharged / removed from the solution (1) M2: Leaving behind \(\text{Na}^+\) and \(\text{OH}^-\) in solution / forming NaOH (1) (d) M1: Platinum is expensive / carbon is much cheaper (1)
題目 4 · structured
10 分
Polymers can be synthesized by different types of polymerization.
(a) Describe two structural differences between addition polymers and condensation polymers. (2)
(b) A polyester is formed by reacting ethane-1,2-diol (\(\text{HO-CH}_2\text{-CH}_2\text{-OH}\)) with butanedioic acid (\(\text{HOOC-CH}_2\text{-CH}_2\text{-COOH}\)).
(i) Draw the structural formula of one repeat unit of this polyester. Show all atoms and covalent bonds in the ester linkage. (3)
(ii) Write a chemical equation for the formation of this polyester using \(n\) molecules of each monomer. (3)
(c) Explain why condensation polyesters are described as biodegradable, whereas addition polymers made from alkenes are not biodegradable. (2)
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解題
(a) Addition polymers are made from monomers with C=C double bonds, resulting in a continuous carbon-only backbone with no other products. Condensation polymers are formed from monomers with functional groups at each end, resulting in a backbone containing ester (or amide) linkages and the release of a small molecule like water. (b)(i) The repeat unit is formed by removing \(\text{-H}\) from each diol group and \(\text{-OH}\) from each carboxylic acid group: \(\text{[-O-CH}_2\text{-CH}_2\text{-O-CO-CH}_2\text{-CH}_2\text{-CO-]}_n\) (b)(ii) \(n\text{ HO-CH}_2\text{-CH}_2\text{-OH} + n\text{ HOOC-CH}_2\text{-CH}_2\text{-COOH} \rightarrow \text{[-O-CH}_2\text{-CH}_2\text{-O-CO-CH}_2\text{-CH}_2\text{-CO-]}_n + 2n\text{ H}_2\text{O}\) (c) Polyesters contain ester linkages which can be broken down (hydrolyzed) by bacteria or other living organisms in the environment. Addition polymers consist of strong, non-polar carbon-carbon single bonds in their backbone, which are inert and cannot be readily attacked or hydrolyzed by microbes.
評分準則
(a) Any two from: M1: Addition polymer backbone has only C-C bonds, condensation backbone has ester/amide linkages (1) M2: Addition polymer is the only product, condensation produces a small molecule / water as well (1) M3: Addition monomer has C=C bond, condensation monomers have -OH, -COOH, or -NH2 groups (1) [Max 2] (b)(i) M1: Correct diol contribution: \(\text{-O-CH}_2\text{-CH}_2\text{-O-}\) (1) M2: Correct acid contribution: \(\text{-CO-CH}_2\text{-CH}_2\text{-CO-}\) (1) M3: Ester linkage shown fully and continuation bonds at each end (1) (b)(ii) M1: Reactants: \(n\text{ HO-CH}_2\text{-CH}_2\text{-OH} + n\text{ HOOC-CH}_2\text{-CH}_2\text{-COOH}\) (1) M2: Correct polymer representation with bracket and subscript n (1) M3: \(+ 2n\text{ H}_2\text{O}\) (1) (c) M1: Ester links in polyesters can be hydrolyzed / broken down by microbes/water (1) M2: Carbon-carbon bonds in addition polymers are strong / inert / cannot be hydrolyzed (1)
題目 5 · structured
10 分
A student investigates the thermal decomposition of a sample of impure magnesium carbonate.
(a) The student heats \(4.50 \text{ g}\) of the impure sample. The carbon dioxide gas is collected and its volume is measured as \(0.960 \text{ dm}^3\) at room temperature and pressure (rtp).
(i) Calculate the amount, in moles, of \(\text{CO}_2\) gas collected. [1 mole of gas occupies \(24.0 \text{ dm}^3\) at rtp] (2)
(ii) Calculate the mass of pure magnesium carbonate (\(M_{\text{r}} = 84.3\)) that decomposed. (2)
(iii) Calculate the percentage purity of the magnesium carbonate sample. (2)
(b) In a separate experiment, the student reacts \(3.00 \text{ g}\) of pure magnesium oxide (\(M_{\text{r}} = 40.3\) ) with \(80.0 \text{ cm}^3\) of \(2.00 \text{ mol/dm}^3\) nitric acid.
Show by calculation which reagent is in excess. (4)
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解題
(a)(i) Moles of \(\text{CO}_2 = \frac{0.960 \text{ dm}^3}{24.0 \text{ dm}^3\text{/mol}} = 0.0400 \text{ mol}\). (a)(ii) The molar ratio of \(\text{MgCO}_3\) to \(\text{CO}_2\) is 1:1. Therefore, moles of \(\text{MgCO}_3\) decomposed = \(0.0400 \text{ mol}\). Mass of \(\text{MgCO}_3 = 0.0400 \times 84.3 = 3.372 \text{ g}\) (or \(3.37 \text{ g}\) to 3 s.f.). (a)(iii) Percentage purity = \(\frac{\text{mass of pure MgCO}_3}{\text{mass of impure sample}} \times 100 = \frac{3.372}{4.50} \times 100 = 74.93\%\) (accept \(74.9\%\) or \(75\%\)). (b) Moles of \(\text{MgO} = \frac{3.00}{40.3} = 0.0744 \text{ mol}\). Moles of \(\text{HNO}_3 = \frac{80.0}{1000} \times 2.00 = 0.160 \text{ mol}\). From the equation, \(1 \text{ mol}\) of \(\text{MgO}\) reacts with \(2 \text{ mol}\) of \(\text{HNO}_3\). Therefore, \(0.0744 \text{ mol}\) of \(\text{MgO}\) requires \(2 \times 0.0744 = 0.1488 \text{ mol}\) of \(\text{HNO}_3\). Since we have \(0.160 \text{ mol}\) of \(\text{HNO}_3\), which is greater than \(0.1488 \text{ mol}\), \(\text{HNO}_3\) is in excess.
評分準則
(a)(i) M1: \(0.960 / 24.0\) (1) M2: \(0.0400\) (mol) (1) (a)(ii) M1: Use of 1:1 molar ratio (1) M2: \(0.0400 \times 84.3 = 3.37\) (g) (1) (a)(iii) M1: \((3.37 / 4.50) \times 100\) (1) M2: \(74.9\%\) or \(75\%\) (1) (b) M1: Moles of \(\text{MgO} = 3.00 / 40.3 = 0.0744\) (mol) (1) M2: Moles of \(\text{HNO}_3 = 0.080 \times 2.0 = 0.160\) (mol) (1) M3: Use stoichiometry ratio of 1:2 (1) M4: State \(\text{HNO}_3\) is in excess with supporting comparison (1)
題目 6 · structured
10 分
Titanium is an important transition metal extracted from its ore, rutile (\(\text{TiO}_2\)), in a multi-stage process.
(a) Explain why titanium cannot be extracted from titanium dioxide by heating with carbon. (2)
(b) In the Kroll process, titanium(IV) chloride (\(\text{TiCl}_4\)) is reacted with liquid sodium at 800 °C:
Explain why this reaction is described as a redox reaction. Identify the oxidizing agent, with reference to electron transfer. (4)
(c) State two precautions needed in this reaction, other than heating, to ensure the reaction proceeds safely and successfully. (2)
(d) Suggest two reasons why the extraction of titanium is much more expensive than the extraction of iron. (2)
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解題
(a) Titanium is more reactive than carbon, so carbon is not a strong enough reducing agent to displace titanium from its oxide. (b) Redox reactions involve both oxidation and reduction occurring simultaneously. Sodium is oxidized because it loses electrons (its oxidation state increases from 0 to +1). Titanium in \(\text{TiCl}_4\) is reduced because it gains electrons (its oxidation state decreases from +4 to 0). \(\text{TiCl}_4\) is the oxidizing agent because it accepts electrons from sodium, causing sodium to be oxidized. (c) The reaction must be carried out under an inert atmosphere (e.g., argon gas) to prevent hot sodium and titanium from reacting with oxygen or water. Also, all reactants and equipment must be completely dry. (d) The process is a batch process (rather than a continuous process like the blast furnace), which takes more time and is less efficient. Also, sodium is expensive to produce (requires electrolysis), and high energy costs are incurred to maintain high temperatures and an argon atmosphere.
評分準則
(a) M1: Titanium is more reactive than carbon (1) M2: Carbon cannot reduce titanium dioxide / cannot displace titanium (1) (b) M1: Both oxidation and reduction occur simultaneously (1) M2: Sodium is oxidized / loses electrons OR titanium ion is reduced / gains electrons (1) M3: \(\text{TiCl}_4\) is the oxidizing agent (1) M4: Because it gains electrons / causes sodium to be oxidized (1) (c) Any two from: M1: Inert atmosphere / use argon gas (1) M2: Exclusion of air/oxygen (1) M3: Exclusion of moisture/water (1) [Max 2] (d) Any two from: M1: Batch process (takes longer/less efficient) (1) M2: Cost of sodium is very high (1) M3: High energy cost of maintaining high temperatures and argon atmosphere (1) [Max 2]
題目 7 · structured
10 分
A student is provided with a green crystalline solid, Compound Y, which is a hydrated transition metal halide.
(a) The student carries out a flame test on Y.
(i) Describe how to carry out a flame test. (3)
(ii) The flame test shows no distinctive color, but adding aqueous sodium hydroxide to a solution of Y produces a green precipitate that is insoluble in excess. State the formula of the cation in Y. (1)
(b) To test for the halide ion, the student adds dilute nitric acid followed by silver nitrate solution to a solution of Y. A cream precipitate forms.
(i) Identify the halide ion present. (1)
(ii) Write an ionic equation, including state symbols, for the formation of this cream precipitate. (2)
(c) Describe a chemical test to show that Compound Y contains water of crystallization. State the observations you would make. (3)
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解題
(a)(i) Clean a platinum or nichrome wire by dipping it in concentrated hydrochloric acid and holding it in a hot non-luminous Bunsen burner flame until no color is observed. Dip the clean wire into the sample of Y, then place it into the non-luminous (blue) Bunsen flame and observe the color produced. (a)(ii) \(\text{Fe}^{2+}\) (Iron(II) ion). (b)(i) Bromide (\(\text{Br}^-\)). (b)(ii) \(\text{Ag}^+\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{AgBr(s)}\) (c) Heat a sample of Y in a test tube. Water droplets will condense at the cooler top of the test tube, and the solid will change color (from green to a different color / anhydrous form). Alternatively, test the liquid produced with anhydrous cobalt(II) chloride paper (turns from blue to pink) or anhydrous copper(II) sulfate (turns from white to blue).
評分準則
(a)(i) M1: Clean platinum/nichrome wire using concentrated hydrochloric acid (1) M2: Dip wire in sample and place in non-luminous / blue flame (1) M3: Observe flame color (1) (a)(ii) M1: \(\text{Fe}^{2+}\) (1) (Accept Iron(II) ion) (b)(i) M1: Bromide / \(\text{Br}^-\right.\) (1) (Reject bromine) (b)(ii) M1: \(\text{Ag}^+ + \text{Br}^- \rightarrow \text{AgBr}\) (1) M2: State symbols \(\text{(aq)}\), \(\text{(aq)}\), \(\text{(s)}\) correct (1) (c) M1: Heat the solid in a test tube (1) M2: Droplets of liquid/water condense on the cooler parts of the tube (1) M3: Solid changes color / turns from green to white/brown (1) OR M1: Add liquid product to anhydrous copper(II) sulfate (1) M2: Turns from white (1) M3: to blue (1)
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